Is ${362985}$ divisible by $3$ ?
A number is divisible by $3$ if the sum of its digits is divisible by $3$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {362985}= &&{3}\cdot100000+ \\&&{6}\cdot10000+ \\&&{2}\cdot1000+ \\&&{9}\cdot100+ \\&&{8}\cdot10+ \\&&{5}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {362985}= &&{3}(99999+1)+ \\&&{6}(9999+1)+ \\&&{2}(999+1)+ \\&&{9}(99+1)+ \\&&{8}(9+1)+ \\&&{5} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {362985}= &&\gray{3\cdot99999}+ \\&&\gray{6\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{9\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {3}+{6}+{2}+{9}+{8}+{5} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $3$ , so the first five terms must all be multiples of $3$ That means that to figure out whether the original number is divisible by $3 $ , all we need to do is add up the digits and see if the sum is divisible by $3$ . In other words, ${362985}$ is divisible by $3$ if ${ 3}+{6}+{2}+{9}+{8}+{5}$ is divisible by $3$ Add the digits of ${362985}$ $ {3}+{6}+{2}+{9}+{8}+{5} = {33} $ If ${33}$ is divisible by $3$ , then ${362985}$ must also be divisible by $3$ ${33}$ is divisible by $3$, therefore ${362985}$ must also be divisible by $3$.